Suppose that X is an angle in standard position and that X' is it's reference angle. If cosX = -4/ squareroot of 17. find the exact value of cosX'
is it 4/square root of 17?
yes, X' would be in quadrant I, so its cosine would be positive.
cos X' = 4/√17
No
To find the exact value of cos(X'), we need to use the relationship between the cosine of an angle and its reference angle.
The reference angle, X', is the acute angle formed between the terminal side of angle X and the x-axis. It is always positive.
Given that cos(X) = -4/√17, we can determine the value of sin(X) using the Pythagorean identity: sin^2(X) + cos^2(X) = 1.
Let's solve for sin(X):
sin^2(X) + cos^2(X) = 1
sin^2(X) + (-4/√17)^2 = 1
sin^2(X) + 16/17 = 1
sin^2(X) = 1 - 16/17
sin^2(X) = 17/17 - 16/17
sin^2(X) = 1/17
sin(X) = ±√(1/17)
Since we are in the standard position, we can determine the sign of sin(X) based on the quadrant the angle X is in. Since cos(X) is negative, X is in either the second or third quadrant.
Now, using the definition of the reference angle, we know that |cos(X)| = cos(X'). Therefore, cos(X') = |cos(X)| = |-4/√17|.
Since √17 is always positive, we can simplify cos(X') as follows:
cos(X') = |-4/√17|
‾‾‾‾‾‾‾‾‾‾‾‾
√17
Therefore, the exact value of cos(X') is 4/√17.